Question: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = \dfrac{1}{2}x + 2}\enspace$ and passes through the point ${(-3, 5)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Solution: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${\dfrac{1}{2}}$ , and its negative reciprocal is ${-2}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -2x + b}\enspace$ We can plug our point, $(-3, 5)$ , into this equation to solve for ${b}$ , the y-intercept. $5 = {-2}(-3) + {b}$ $5 = 6 + {b}$ $5 - 6 = {b} = -1$ The equation of the perpendicular line is $\enspace {y = -2x - 1}\enspace$. ${m = -2, \enspace b = -1}$